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Let $\mathbf{Inv}\left(\mathfrak{gl}(r,\mathbb{R})\right)$ be the set of invariant polynomials on $\mathfrak{gl}(r,\mathbb{R})$, i.e $P\in\mathbf{Inv}\left(\mathfrak{gl}(r,\mathbb{R})\right)$ iff: $$ P(AXA^{-1})=P(X) \quad \forall A \in GL(r,\mathbb{R}) $$ For example, trace polynomials $\Sigma_k(X)=\text{tr}(X^k)$ are invariant polynomials for all $k$. Also for $\det(I+\lambda X)=\sum_{k=0}^{r}\lambda^kf_k(X)$, the polynomials $f_k(X)$ are invariant polynomials.

The theorem I would like a prove is that $\mathbf{Inv}\left(\mathfrak{gl}(r,\mathbb{R})\right)$ is generated as a ring by $\Sigma_k$, and also by $f_k$: $$ \mathbf{Inv}\left(\mathfrak{gl}(r,\mathbb{R})\right)=\mathbb{R}\left[\Sigma_1,\dots,\Sigma_r\right]=\mathbb{R}\left[f_1,\dots,f_r\right] $$

I've read this theorem in many places, but none provided a proof. It's Theorem 23.4 in Loring Tu "Differential Geometry: Connections, Curvature, and Characteristic Classes".

Diana Pestana
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  • It seems that if you consider invertible diagonal matrices $\mathrm{diag}(a_1,\ldots,a_r)$ you obtain a symmetric polynomial in $a_1,\ldots,a_r$, and the algebra of symmetric polynomials is generated by elementary symmetric polynomials – AAAL Dec 01 '24 at 15:06
  • The result is true over any field of characteristic zero, see https://math.stackexchange.com/questions/181237/what-is-the-structure-of-invariant-matrix-polynomials/181252#181252 for a discussion. – Qiaochu Yuan Dec 01 '24 at 16:17

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